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Problem V.

To find the diameter and circumference of a circle, either from the other.

Rule 1. As 7 is to 22, so is the diameter to the circumference, anri as 22 is to 7, so is the circumference to the diameter.

Rule 2. As 113 is to 355, so is the diameter to the circumference, and as 355 isito 113, so is the circumference to the diameter.

Rule 3. As 1 is to 3.1416, so is the diameter to the circuinference, and as 3.1416 is to 1, so is the circumference to the diameter:

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Examples. 1. What is the circumference 3. What is the diameter of a.. of a circle whose diameter is 14ft. circle whose circumference is 50: By Rule 1.

rods ?

By Rule 1.
As 7:22 :: 14:44 Ans.

As 22:7 ::50:15.9090 Ans.
By Rule 2.

By Rule 2.
As 113:355 :: 14:4311} Ans.

As 355:113::50:15.9155 Ans,
By Rule 3.

By Rule 3.
As 1: 3.1416:: 14:43.9824 Ans.* | As 3,1416:1::50:15.9156 Ans.

2. Supposing the diameter of 4. Supposing the circumferthe earth to be 7958 miles, what ence of the earth to be 25000 is its circumference ?

miles, what is its diameter ? Ans. 25000.8528 miles..

Ans. 7957| nearly.

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Problem VI.

To find the area of a circle.t. RULE.--Multiply half the circumference by half the diameter, and the product will be the area.

* These three methods do not exactly agree, but the last is the most correct.
The exact proportion between the diameter and circumference of a circle has
not yet been ascertained.
+ The following are some of the most useful problems relating to the citcles

1. Circumference x diameter, the product=the area, a cijev 1933
2. Square of diameter * .7854=area.
3. Square of circumference'x .07958=area.
4. As 14:11 :: square of diameter : area.
5. As 88 : 7 :: square of circumference :-area.
6. Diameter x .886l=side of an equal square.
7. Circumference x 2821=side of an equal square.
8. Diameter x .7071=side of an inscribed-square.

..IJUSI 9. Circumference .2251=side of an inscribed square.

o bob 10. Area *.6366—side of an inscribed square. 11. Side of a square x 1.128—diameter of an equal circle. 12. Side of a square 3.545=circumference of an equal crole

901

Examples. 1. What is the area of a circle 3. What is the area of a circle whose diameter is 7 and circum whose diameter is 10 rods, and ference 22 feet?

circumference 31.416? 11=į circumference.

Ans. 78.54 rods. 3.5= 1 diameter.

4. How many square

chains in 55

a circular field, whose circumfe33

rence is 44 chains, and diameter 14?

Ans. 154 chains. 38.5 feet. Ans. 2. What is the area of a circle 5. How many square feet in a whose diaineter is 1. and circum circle whose circumference is 63 ference 3.1416? Ans. .7854 feet?

Ans. 315 feet.

Problem VII. The area of a circle given to find the diameter and circumference. Rule.-1. Divide the area by .7854, and the square root of the quotient will be the diameter.

2. Divide the area by .07958, and the square root of the quotient will be the circumference.

Examples. 1. What is the diameter of a 3. I demand the length of a circle whose area is 154 rods ? rope to be tied to a horse's neck,

that he may graze upon 7854 .7854)154.0000(196(14 rods, Ans.

square feet of new feed every 7854 1

day for 4 days, one end of the 75460 24)96

rope being each day fastened to 70686

the same stake? 96

'Ist circle contains 7854 feet 47740

-.7854= 10000, and 10000= 47124

100 diam. :-2=50 feet, the 1st

rope. 2d circle contains 15708 616

7854=20000, and 720000= 1414 704 feet, second

rope,

&c. 2. The area of a circle is 78.5

1st rope 50 feet.

2 701 feet. feet; what is its circumference ?

Ans. 3

86į feet Ans. 31.4 feet.

4

100 feet.

Problem VIII.

To find the area of an oval, or ellipsis. Rule.-Multiply the longest and shortest diameters together, and the product by 7854; the last product will be the area.*

* The Longest diameter of an ellipsiş is called the transverse, and the shortest the conjugate diameter.

Examples 1. What is the area of an oval 2. What is the area of an oval whose longest diameter is 5 feet, whose longest diameter is 21, and and shortest 4 feet?

shortest 17 ? 5X4X.7854= 15.708 ft. Ans.

Ans. 280.3878. Problem IX.

To find the area of a globe or sphere. Rule.-Multiply the circumference by the diameter, and the product will be the area.

Examples.' 1. How many square feet in 3. What is the area of the surthe surface of a globe whose di- | face of a cannon shot, whose diameter is 14 inches and circuin- ameter is one inch ? ference 44 ?

Ans. 3.1416 inches. 44 x 14=616 Ans. 2. How many square miles in

4. How many square inches in

the surface of an 18 inch artifithe earth's surface, its circumfer

cial globe ? Ans. 1017.8784. ence being 25000, and its diameter 7957% iniles :

Ans. 198943750.

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2. Fensuration of Solids.

Definitions. 1. A solid is a figure having 5. A cylinder is a round prism, three dimensions, viz. length, having circles for its ends. breadth and thickness.

6. A pyramid is a solid whose 2. A prism is a body whose base is any plane figure, and ends are any equal and similar whose sides are triangular, meetplane figures, and whose sides ing in a point at the top called a are parallelograms.

vertex. 3. A cube is a body having six 7. A cone is a round pyramid, equal sides, all of which are having a circle for its base. squares.

8. A sphere is a solid bounded 4. A parallelopipedon is a body by one continued convex surface, having six rectangular sides, ev. every part of which is equally ery opposite pair of which are i distant from a point within called equal and parallel.

the centre. Mensuration of Solids teaches to determine the spaces included by contiguous surfaces, and the sum of the measures of these including surfaces is the whole surface of the body. The measure of a solid is called its solidity, capacity, or content. The content is estimated by the number of cubes, whose sides are inches, or feet, or yards, &c. contained in the body.

answer.

dity

Problem I.

To find the solidity of a cube. Rule-Cube one of its sides, that is, multiply the side by itself, and that product by the side again, and the last product will be the

Examples. 1. If the length of the side of a 2. How many cubick inches in cube be 22 feet, what is its soli. a cube whose side is 24 inches ?

Ans. 13824. 22 x 22 x 22=10648 Ans.

Problem II.

To find the solidity of a parallelopipedon. RULE.—Multiply the length by the breadth, and that product by the depth, the last product will be the answer.

Examples. 1. What is the content of a 2. How many feet in a stick parallelopipedon whose length is of hewn timber 30 feet long, 9 6 feet, its breadth 24 feet, and inches broad, and 6 inches thick? its depth 14 feet?

Ans. 114 feet. 6 x 2.5 x 1.75=26.25 or 26 feet.

Problem III. To find the side of the largest square stick of timber that can be hewn from a

round log: RULE.—Extract the square root of twice the square of the semidiameter at the smallest end for the side of the stick when squared.

Examples 1. The diameter of a round 2. The diameter at the smalllog at its smallest end is 16 in- est end being 24 inches, how ches, what will be the side of the large square will the stick of largest squared stick of timber timber hew? Ans. 16.97 in. that can be hewn from it?

78X8X2=11.31 in. Ans.

Problem IV.

To find the solidity of a prism, or cylinder. Rule.-Multiply the area of the end by the length of the prism, for the content.

Examples 1. What is the content of a 2. What number of cubick feet triangular prism, the area of in a round stick of timber, whose whose end is 2.7 feet, and whose diameter is 18 inches, and length length is 12 feet?

20 feet?

Ans. 35.343. 2.7 X 12=32.4 ft. Ans.

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Problem V.

To find the solidity of a pyramid or cone, Rule.--Multiply the area of the base by the height, and one third of the product will be the content.

Examples. 1. What is the content of a 2. What is the content of a cone whose height is 124 feet, and I triangular pyramid, its height bethe diameter of the base 24 feet ? | ing 144 feet, and the sides of its

24 x 21= x=2=61 and base being 5, 6, and 7 feet? 61 X.7854 x 122:3=20.453125.

Ans. 71.035+

Problem VI.

To find the solidity of a sphere.* RULE.--Multiply the cube of the diameter by .5236, or multiply the square of the diameter by f the circumference.

Examples. 1. What is the content of a 2. What is the solid content sphere whose diameter is 12 in- of the earth, its circumference ches ? 12 x 12 x 12 x 5236= being 25000 miles ? 904.7808 Ans.

Ans. 263858149120 miles.

3. Gauging.

Gauging teaches to measure all kinds of vessels, as pipes, hogsheads, barrels, &c.

RuLÈ.- To the square of the bung diameter add the square of the head diameter; multiply the sum by the length, and the product by .0014 for ale gallons, or by .0017 for wine gallons.

Examples. 1. What is the content of a 2. What is the content of a cask whose length is 40 inches, cask whose length is 20 inches, and its diameters 24 and 32 in- | and diameters 12 and 16 ? ches?

11.2 a. gal. Ans.

13.6 w.gal. 32 x 32+ 24 x 24 x 40=64000A. 64000 X 0014–89.6 a.gal.

A.

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* The surface of a sphere is found by multiplying its diameter by its circum. ference.

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